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Existence of Ground State Solutions and Infinitely Many Solutions for a Class of Liquid Crystal Systems
DOI: 10.12677/AAM.2022.1111863, PDF , HTML, XML, 下载: 118  浏览: 200  科研立项经费支持

Abstract: In this paper, we aim to prove the existence of ground state solutions for a class of liquid crystal system. Under the assumptions of V and the nonlinearity g, we find this solution using the Nehari manifold. After that, we prove the existence of nontrivial solutions of liquid crystal system by using the method of Mountain Pass Theorem. Finally, we have made some improvements, and prove a different type of multiplicity result by applying the Krasnoselskii genus theory, the existence of infinitely many solutions to the system.

1. 引言

$\left\{\begin{array}{ll}-\Delta u+V\left(x\right)u+\gamma u\theta =g\left(u\right),\hfill & x\in {ℝ}^{2}.\hfill \\ -\Delta \theta +{c}^{2}={u}^{2},\hfill & x\in {ℝ}^{2},\hfill \end{array}$ (1)

$\left\{\begin{array}{l}2i\frac{\partial E}{\partial z}+{\Delta }_{x,y}E+\alpha \left[{\mathrm{sin}}^{2}\theta -{\mathrm{sin}}^{2}{\theta }_{0}\right]E=0,\\ 2{\Delta }_{x,y}\theta +\left[\beta +\alpha {|E|}^{2}\right]\mathrm{sin}\left(2\theta \right)=0,\end{array}$ (2)

$\theta =\stackrel{¯}{\theta }+{\theta }_{0}$，其中， $\stackrel{¯}{\theta }$ 对应于光诱导的分子重定向，并且 $\stackrel{¯}{\theta }\ll 1$$\stackrel{¯}{\alpha }=\alpha \mathrm{sin}\left(2{\theta }_{0}\right)$$\stackrel{¯}{\beta }=2\beta \mathrm{cos}\left(2{\theta }_{0}\right)$。利用一阶近似，可推得下列的低阶近似模型：

$\left\{\begin{array}{l}2i\frac{\partial E}{\partial z}+{\Delta }_{x,y}E+\stackrel{¯}{\alpha }\stackrel{¯}{\theta }E=0,\\ 2{\Delta }_{x,y}\stackrel{¯}{\theta }+\stackrel{¯}{\beta }\stackrel{¯}{\theta }+\stackrel{¯}{\alpha }{|E|}^{2}=0.\end{array}$ (3)

$\left\{\begin{array}{l}2i\frac{\partial E}{\partial z}+{\Delta }_{x,y}E+\gamma \psi E=0,\\ {\Delta }_{x,y}\psi -{c}^{2}\psi =4\pi {|E|}^{2}=0,\end{array}$ (4)

$E\left(x,y,z\right)={\text{e}}^{i\omega z}u\left(x,y\right)$ ,

$\left\{\begin{array}{ll}\Delta u-2\omega u+\gamma u\psi =0,\hfill & \left(x,y\right)\in {ℝ}^{2}.\hfill \\ -\Delta \psi +{c}^{2}\psi =4\pi {u}^{2},\hfill & \left(x,y\right)\in {ℝ}^{2}.\hfill \end{array}$ (5)

$c=0$ 时，(5)退化为下列Schrödinger-Possion系统

$\left\{\begin{array}{ll}-\Delta u+V\left(x\right)u+\varphi u=f\left(x,u\right),\hfill & x\in {ℝ}^{2},\hfill \\ \Delta \varphi ={u}^{2},\hfill & x\in {ℝ}^{2}.\hfill \end{array}$ (6)

(V1) $V\in C\left({ℝ}^{2},\left(0,\infty \right)\right)$，存在 ${\alpha }_{0}>0$，使得 $\underset{{ℝ}^{2}}{\mathrm{inf}}V\left(x\right)\ge {\alpha }_{0}>0$，并且有 $\underset{|x|\to \infty }{\mathrm{lim}}V\left(x\right)=+\infty$

(g1) $g\in {C}^{1}\left(ℝ,ℝ\right)$，存在 ${\alpha }_{0}>0$，使得

(i)当 $\alpha >{\alpha }_{0}$ 时， $\underset{t\to +\infty }{\mathrm{lim}}\frac{g\left(t\right)}{{\text{e}}^{\alpha {t}^{2}}}=0$

(ii) 当 $\alpha <{\alpha }_{0}$ 时， $\underset{t\to +\infty }{\mathrm{lim}}\frac{g\left(t\right)}{{\text{e}}^{\alpha {t}^{2}}}=+\infty$

(g2) $\underset{t\to 0}{\mathrm{lim}}\frac{g\left(t\right)}{t}=0$

(g3) 存在 $\vartheta >2$，使得对任意的 $t>0$，都有 $0<\vartheta G\left(t\right)\le g\left(t\right)t$

(g4) 函数 $t\to \frac{g\left(t\right)}{{t}^{3}}$ 在区间 $\left(0,+\infty \right)$ 上单调递增；

(g5) 存在 $p>4$$\tau >\mathrm{max}\left\{1,{\tau }^{*}\right\}$，使得 $g\left(t\right)>\tau {t}^{p-1},\forall t>0$，其中，

${\tau }^{*}:={\left(\frac{2p{\alpha }_{0}{c}_{p}}{\pi \left(p-4\right)}\right)}^{\frac{p-2}{2}}$ .

$\left\{\begin{array}{ll}-\Delta u+V\left(x\right)u-\gamma u\theta =\gamma g\left(u\right),\hfill & x\in {ℝ}^{2}.\hfill \\ -\Delta \theta +{c}^{2}\theta ={u}^{2},\hfill & x\in {ℝ}^{2}\hfill \end{array}$

2. 准备性工作

${H}^{1}\left({ℝ}^{2}\right)=\left\{u\in {L}^{2}\left({ℝ}^{2}\right)|\nabla u\in {L}^{2}\left({ℝ}^{2}\right)\right\}$，其上赋予范数

$‖u‖={\left(\underset{{ℝ}^{2}}{\int }\left({|u|}^{2}+{|\nabla u|}^{2}\right)\text{d}x\right)}^{\frac{1}{2}}$ .

$E:=\left\{u\in {H}^{1}\left({ℝ}^{2}\right)|\underset{{ℝ}^{2}}{\int }V\left(x\right){u}^{2}\text{d}x<\infty \right\}$ ,

${‖u‖}_{E}=\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+V\left(x\right){u}^{2}\right)\text{d}x$

(i) $\theta$ 是系统(1)第二个方程的解；

(ii) $\theta$ 满足： $‖\theta \left(u\right)‖\le C{‖u‖}_{p}^{2}$，其中 $p>2$，C是一个常数；

(iii) $\theta :{H}^{1}\left({ℝ}^{2}\right)\to {H}^{1}\left({ℝ}^{2}\right)$ 连续；

(iv) 在 ${ℝ}^{2}$ 上， $\theta \left(u\right)\ge 0$，且对于任意的 $s\in ℝ$$\theta \left[su\right]={s}^{2}\theta \left[u\right]$

$\theta \left[u\right]\left(x,y\right)=\underset{{ℝ}^{2}}{\int }4\pi {E}_{2}\left(c\left(x-s\right),c\left(y-t\right)\right){u}^{2}\left(s,t\right)\text{d}t$ ,

(i) ${K}_{0}\in {C}^{\infty }\left(\left(0,\infty \right)\right)$${K}_{0}\left(r\right)>0$$r>0$，并且 ${\left\{{K}_{0}\left(r\right)\right\}}^{\prime }=-{K}_{1}\left(r\right)$$r>0$

(ii) 当 $r\to 0$ 时，有 $\frac{{K}_{0}\left(r\right)}{-\mathrm{ln}r}\to 1$$r{{K}^{\prime }}_{0}\left(r\right)\to -{L}_{0}$${L}_{0}>0$

(iii) 当 $r\to \infty$ 时， ${\text{e}}^{r}{K}_{0}\left(r\right)\to 0$${\text{e}}^{r}{{K}^{\prime }}_{0}\left(r\right)\to 0$

$-\Delta u+V\left(x\right)u+\gamma u\theta =g\left(u\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in {ℝ}^{2}$ (7)

$I\left(u\right)=\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+V\left(x\right){u}^{2}\right)\text{d}x+\frac{\gamma }{4}\underset{{ℝ}_{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-\underset{{ℝ}_{2}}{\int }G\left(u\right)\text{d}x$ (8)

$〈{I}^{\prime }\left(u\right),v〉=\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left(\nabla u\nabla v+V\left(x\right)uv\right)\text{d}x+\frac{\gamma }{4}\underset{{ℝ}_{2}}{\int }u\theta \left[u\right]v\text{d}x-\underset{{ℝ}_{2}}{\int }g\left(u\right)v\text{d}x$

$\underset{n\to \infty }{\mathrm{lim}}\underset{{ℝ}^{2}}{\int }\theta \left({u}_{n}\right){u}_{n}^{2}\text{d}x=\underset{{ℝ}^{2}}{\int }\theta \left(u\right){u}^{2}\text{d}x$ (9)

$\underset{{ℝ}^{2}}{\int }\left({e}^{\alpha {|u|}^{2}}-1\right)\text{d}x\le C$ .

$|g\left(t\right)|\le \epsilon |t|,\text{\hspace{0.17em}}\text{\hspace{0.17em}}|G\left(t\right)|\le \frac{1}{2}\epsilon {|t|}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}0 (10)

$|g\left(t\right)|\le \epsilon \left({e}^{\alpha {t}^{2}}-1\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\ge {\delta }_{1}$ .

$|g\left(t\right)t|\le \frac{\epsilon }{{\delta }_{1}^{2}}{t}^{2}\left({e}^{\alpha {t}^{2}}-1\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}|G\left(t\right)|\le \frac{\epsilon }{{\delta }_{1}^{2}}\left({e}^{\alpha {t}^{2}}-1\right)$ (11)

$\underset{{ℝ}^{2}}{\int }g\left(u\right)u\text{d}x\le \epsilon \underset{{ℝ}^{2}}{\int }{|u|}^{2}\text{d}x+{C}_{\epsilon }\underset{{ℝ}^{2}}{\int }{|u|}^{q}\left({\text{e}}^{\alpha {u}^{2}}-1\right)\text{d}x$ (12)

$\underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x\le \frac{\epsilon }{2}\underset{{ℝ}^{2}}{\int }{|u|}^{2}\text{d}x+{\stackrel{¯}{C}}_{\epsilon }\underset{{ℝ}^{2}}{\int }{|u|}^{q}\left({\text{e}}^{\alpha {u}^{2}}-1\right)\text{d}x$ (13)

$\begin{array}{c}h\left(s\right)=g\left(s\right)s-4G\left(s\right)=\frac{g\left(s\right)}{{s}^{3}}{s}^{4}-4G\left(t\right)+4{\int }_{s}^{t}g\left(\tau \right)\text{d}\tau \\ \le \frac{g\left(t\right)}{{t}^{3}}{s}^{4}-4G\left(t\right)+\frac{g\left(t\right)}{{t}^{3}}\left({t}^{4}-{s}^{4}\right)

3. 定理1.1 的证明

$N=\left\{u\in E\\left\{0\right\},〈{I}^{\prime }\left(u\right),u〉=0\right\}$ .

$\begin{array}{c}{{\eta }^{\prime }}_{u}\left(t\right)=t\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+V\left(x\right){u}^{2}\right)\text{d}x+\gamma {t}^{3}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-\underset{{ℝ}^{2}}{\int }g\left(tu\right)u\text{d}x\\ \ge t{‖u‖}_{E}^{2}+\gamma {t}^{3}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-\epsilon t\underset{{ℝ}^{2}}{\int }{|u|}^{2}\text{d}x-{t}^{q-1}{\stackrel{¯}{C}}_{\epsilon }\underset{{ℝ}^{2}}{\int }{|u|}^{q}\left({\text{e}}^{\alpha {|u|}^{2}}-1\right)\text{d}x\\ \ge t\left(\frac{1}{2}-\frac{C\epsilon }{2}\right){‖u‖}_{E}^{2}+\gamma {t}^{3}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-{t}^{q-1}{\stackrel{¯}{C}}_{\epsilon }{\left(\underset{{ℝ}^{2}}{\int }{|u|}^{q{s}^{\prime }}\text{d}x\right)}^{\frac{1}{{s}^{\prime }}}{\left[\underset{{ℝ}^{2}}{\int }\left({\text{e}}^{\alpha s{‖tu‖}_{{H}^{1}}^{2}{\left(\frac{u}{{‖u‖}_{{H}^{1}}}\right)}^{2}}-1\right)\text{d}x\right]}^{\frac{1}{s}}\end{array}$ ,

${\eta }_{u}\left(t\right)\ge {D}_{1}t+{D}_{2}{t}^{3}-{D}_{3}{t}^{q-1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}0

${{\eta }^{\prime }}_{u}\left(t\right)\le t{‖u‖}_{E}^{2}+\gamma {t}^{3}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-\tau {t}^{p-1}\underset{{ℝ}^{2}}{\int }{u}^{p}\text{d}x$ .

${u}_{n}\to 0$ (14)

${‖u‖}_{E}^{2}\le \gamma \underset{{ℝ}^{2}}{\int }{u}_{n}^{2}\theta \left[{u}_{n}\right]\text{d}x+\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right){u}_{n}\text{d}x\le \epsilon \underset{{ℝ}^{2}}{\int }{|{u}_{n}|}^{2}\text{d}x+{C}_{\epsilon }\underset{{ℝ}^{2}}{\int }{|{u}_{n}|}^{q}\left({e}^{\alpha {u}_{n}^{2}}-1\right)\text{d}x$

$\left(1-{C}_{\epsilon }\right){‖{u}_{n}‖}_{E}^{2}\le {\stackrel{¯}{C}}_{\epsilon }{\left(\underset{{ℝ}^{2}}{\int }{|{u}_{n}|}^{q{s}^{\prime }}\text{d}x\right)}^{\frac{1}{{s}^{\prime }}}{\left[\underset{{ℝ}^{2}}{\int }\left({e}^{\alpha s{‖{u}_{n}‖}_{{H}^{1}}^{2}{\left(\frac{u}{{‖{u}_{n}‖}_{{H}^{1}}}\right)}^{2}}-1\right)\text{d}x\right]}^{\frac{1}{s}}$

$\left(1-{C}_{\epsilon }\right){‖{u}_{n}‖}_{E}^{2}\le M{C}_{\epsilon }{\left(\underset{{ℝ}^{2}}{\int }{|{u}_{n}|}^{q{s}^{\prime }}\text{d}x\right)}^{\frac{1}{{s}^{\prime }}}\le M{C}_{\epsilon }C{‖{u}_{n}‖}_{E}^{q}$ .

$\underset{u\in N}{\mathrm{inf}}I\left(u\right)=c$

$\begin{array}{c}I\left(u\right)=I\left(u\right)-\frac{1}{4}〈{I}^{\prime }\left(u\right),u〉\\ =\frac{1}{4}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+V\left(x\right){u}^{2}\right)\text{d}x+\frac{1}{4}\underset{{ℝ}_{2}}{\int }\left(\frac{1}{4}g\left(u\right)u-G\left(u\right)\right)\text{d}x\\ \ge \frac{1}{4}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+V\left(x\right){u}^{2}\right)\text{d}x\ge C>0\end{array}$ .

$c+{o}_{n}\left(1\right)=I\left({u}_{n}\right)-\frac{1}{4}〈{I}^{\prime }\left({u}_{n}\right),{u}_{n}〉=\frac{1}{4}{‖{u}_{n}‖}_{E}^{2}+\underset{{ℝ}^{2}}{\int }\left(\frac{1}{4}g\left({u}_{n}\right){u}_{n}-G\left({u}_{n}\right)\right)\text{d}x\ge \frac{1}{4}{‖{u}_{n}‖}_{E}^{2}$

$-\Delta u+V\left(x\right)u+\gamma u\theta ={|u|}^{p-2}u,\text{}{ℝ}^{2},$

${I}_{p}\left(u\right)=\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+V\left(x\right){u}^{2}\right)\text{d}x+\frac{\gamma }{4}\underset{{ℝ}_{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-\frac{1}{p}\underset{{ℝ}_{2}}{\int }{|u|}^{p}\text{d}x$ .

${N}_{p}=\left\{u\in E\\left\{0\right\},〈{{I}^{\prime }}_{p}\left(u\right),u〉=0\right\}$

${c}_{p}={I}_{p}\left({w}_{p}\right)={I}_{p}\left({w}_{p}\right)-4〈{{I}^{\prime }}_{p}\left({w}_{p}\right),{w}_{p}〉=\frac{1}{4}{‖{w}_{p}‖}_{E}^{2}+\left(\frac{1}{4}-\frac{1}{p}\right)\underset{{ℝ}^{2}}{\int }{w}_{p}^{p}\text{d}x\ge \frac{p-4}{4p}\underset{{ℝ}^{2}}{\int }{w}_{p}^{p}\text{d}x$ (15)

${‖{w}_{p}‖}_{E}^{2}+\gamma \underset{{ℝ}^{2}}{\int }{w}_{p}^{2}\theta \left[{w}_{p}\right]\text{d}x=\underset{{ℝ}^{2}}{\int }{w}_{p}^{p}\text{d}x\le \tau \underset{{ℝ}^{2}}{\int }{w}_{p}^{p}\text{d}x\le \underset{{ℝ}^{2}}{\int }g\left({w}_{p}\right){w}_{p}\text{d}x$

$\begin{array}{c}c\le \frac{{\beta }^{2}}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla {w}_{p}|}^{2}+V\left(x\right){w}_{p}^{2}\right)\text{d}x+\frac{\gamma {\beta }^{4}}{4}\underset{{ℝ}_{2}}{\int }{w}_{p}^{2}\theta \left[{w}_{p}\right]\text{d}x-\underset{{ℝ}_{2}}{\int }G\left(\beta {w}_{p}\right)\text{d}x\\ \le \frac{{\beta }^{2}}{2}\left({‖{w}_{p}‖}_{E}^{2}+\frac{\gamma }{2}\underset{{ℝ}_{2}}{\int }{w}_{p}^{2}\theta \left[{w}_{p}\right]\text{d}x\right)-\frac{\tau }{p}{\beta }^{p}\underset{{ℝ}_{2}}{\int }{w}_{p}^{p}\text{d}x\\ \le \frac{{\beta }^{2}}{2}\left({‖{w}_{p}‖}_{E}^{2}+\gamma \underset{{ℝ}_{2}}{\int }{w}_{p}^{2}\theta \left[{w}_{p}\right]\text{d}x\right)-\frac{\tau }{p}{\beta }^{p}\underset{{ℝ}_{2}}{\int }{w}_{p}^{p}\text{d}x\\ \le \left(\frac{{\beta }^{2}}{2}-\frac{\tau }{p}{\beta }^{p}\right)\underset{{ℝ}_{2}}{\int }{w}_{p}^{p}\text{d}x\end{array}$ .

$c\le \frac{1}{2}{\left(\frac{1}{\tau }\right)}^{\frac{2}{p-2}}\frac{4p{c}_{p}}{p-4}=\frac{2p{c}_{p}}{\left(p-4\right){\tau }^{\frac{2}{p-2}}}$ .

$〈{I}^{\prime }\left({u}_{n}\right),{u}_{n}〉=0$，可推得 ${‖{u}_{n}‖}_{E}\to 0$，这与引理3.2矛盾，得证。

$\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right){u}_{n}\text{d}x\to \underset{{ℝ}^{2}}{\int }g\left(u\right)u\text{d}x$

$\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right)w\text{d}x\to \underset{{ℝ}^{2}}{\int }g\left(u\right)w\text{d}x,\text{}\forall w\in E$ .

${‖{u}_{n}‖}_{E}^{2}\le \underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\left({‖\nabla {u}_{n}‖}_{{L}^{2}}^{2}+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}_{n}^{2}\text{d}x\right)$

${‖{u}_{n}‖}_{E}^{2}\le \gamma \underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x+\underset{{ℝ}^{2}}{\int }g\left(u\right)u\text{d}x$ .

$c\le I\left(tu\right)=I\left(tu\right)-\frac{1}{4}〈{I}^{\prime }\left(tu\right),tu〉=\frac{1}{4}{‖tu‖}_{E}^{2}+\underset{{ℝ}^{2}}{\int }\left(\frac{1}{4}g\left(tu\right)tu-4G\left(tu\right)\right)\text{d}x$ .

$0，则利用引理2.4和Fatou's引理，得 $c\le I\left(tu\right)<\frac{1}{4}{‖u‖}_{E}^{2}+\underset{{ℝ}^{2}}{\int }\left(\frac{1}{4}g\left(u\right)u-4G\left(u\right)\right)\text{d}x\le \frac{1}{4}\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}\left({‖{u}_{n}‖}_{E}^{2}+\underset{{ℝ}^{2}}{\int }\left(g\left(u\right)u-4G\left(u\right)\right)\text{d}x\right)$$c\le I\left(tu\right)<\underset{n\to \infty }{\mathrm{lim}}I\left({u}_{n}\right)=c$，说明 $0 不可能发生。因此 $t=1$，即 $u\in N$$I\left(u\right)=c$，得证。

4. 定理1.2的证明注意到 $I\left(0\right)=0$，由引理2.1，以及(12)可得，存在常数 $C>0$，使得

$\begin{array}{c}I\left(u\right)=\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+V\left(x\right){u}^{2}\right)\text{d}x+\frac{\gamma }{4}\underset{{ℝ}_{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-\underset{{ℝ}_{2}}{\int }G\left(u\right)\text{d}x\\ \ge \frac{1}{2}{‖u‖}_{E}^{2}+{C}_{1}\gamma {‖u‖}_{E}^{4}-\frac{\epsilon }{2}\underset{{ℝ}^{2}}{\int }{|u|}^{2}\text{d}x-{\stackrel{¯}{C}}_{\epsilon }\underset{{ℝ}^{2}}{\int }{|u|}^{q}\left({\text{e}}^{\alpha {u}^{2}}-1\right)\text{d}x\end{array}$

$\underset{{ℝ}^{2}}{\int }\left({|u|}^{q}\left({\text{e}}^{\alpha {u}^{2}}-1\right)\right)\text{d}x\le {C}_{\epsilon }{\left(\underset{{ℝ}^{2}}{\int }{|u|}^{q{s}^{\prime }}\text{d}x\right)}^{\frac{1}{{s}^{\prime }}}{\left[\underset{{ℝ}^{2}}{\int }\left({\text{e}}^{\alpha s{‖{u}_{n}‖}_{{H}^{1}}^{2}{\left(\frac{u}{{‖{u}_{n}‖}_{{H}^{1}}}\right)}^{2}}-1\right)\text{d}x\right]}^{\frac{1}{s}}\le C{‖u‖}_{q{s}^{\prime }}^{q}\le C{‖u‖}_{E}^{q}$ (16)

$I\left(u\right)\ge \left(\frac{1}{2}-\frac{\epsilon }{2}\right){‖u‖}_{E}^{2}+\gamma {D}_{1}{‖u‖}_{E}^{4}-{D}_{2}{‖u‖}_{E}^{q}$ .

$I\left(s{u}_{0}\right)=\frac{{s}^{2}}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla {u}_{0}|}^{2}+V\left(x\right){u}_{0}^{2}\right)\text{d}x+\frac{\gamma {s}^{4}}{4}\underset{{ℝ}_{2}}{\int }{u}_{0}^{2}\theta \left[{u}_{0}\right]\text{d}x-\underset{{ℝ}_{2}}{\int }G\left(s{u}_{0}\right)\text{d}x,\text{}s>0$ .

$I\left({u}_{n}\right)\to d,\text{}{I}^{\prime }\left({u}_{n}\right)\to 0$

$\underset{n\to \infty }{\mathrm{lim}}\left[\underset{{ℝ}^{2}}{\int }\left({|\nabla {u}_{n}|}^{2}+V\left(x\right){u}_{n}^{2}\right)\text{d}x-4\underset{{ℝ}_{2}}{\int }G\left({u}_{n}\right)\text{d}x+\underset{{ℝ}_{2}}{\int }g\left({u}_{n}\right){u}_{n}\text{d}x\right]=\underset{n\to \infty }{\mathrm{lim}}\left[4I\left({u}_{n}\right)-〈{I}^{\prime }\left({u}_{n}\right),{u}_{n}〉\right]=4d$

$\underset{{ℝ}_{2}}{\int }g\left({u}_{n}\right){u}_{n}\text{d}x-4\underset{{ℝ}_{2}}{\int }G\left({u}_{n}\right)\text{d}x\ge 0$ .

$\begin{array}{l}\underset{{ℝ}^{2}}{\int }\theta \left[{u}_{n}\right]{u}_{n}\phi \text{d}x-\underset{{ℝ}^{2}}{\int }\theta \left[u\right]u\phi \text{d}x\\ =\underset{{ℝ}^{2}}{\int }\left(\theta \left[{u}_{n}\right]-\theta \left[u\right]\right){u}_{n}\phi \text{d}x+\underset{{ℝ}^{2}}{\int }\theta \left[u\right]\left({u}_{n}-u\right)\phi \text{d}x\\ \le {\left(\underset{{ℝ}^{2}}{\int }{\left(\theta \left[{u}_{n}\right]-\theta \left[u\right]\right)}^{2}\text{d}x\right)}^{\frac{1}{2}}{\left(\underset{{ℝ}^{2}}{\int }{|{u}_{n}|}^{3}\text{d}x\right)}^{\frac{1}{3}}{\left(\underset{{ℝ}^{2}}{\int }{|\phi |}^{6}\text{d}x\right)}^{\frac{1}{6}}\\ \text{}+{\left(\underset{{ℝ}^{2}}{\int }{\left(\theta \left[{u}_{n}\right]-\theta \left[u\right]\right)}^{2}\text{d}x\right)}^{\frac{1}{2}}{\left(\underset{{ℝ}^{2}}{\int }{|{u}_{n}-u|}^{3}\text{d}x\right)}^{\frac{1}{3}}{\left(\underset{{ℝ}^{2}}{\int }{|\phi |}^{6}\text{d}x\right)}^{\frac{1}{6}}\to 0\end{array}$ (17)

$\underset{{ℝ}^{2}}{\int }\nabla {u}_{n}\nabla \phi \text{d}x+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}_{n}\phi \text{d}x\to \underset{{ℝ}^{2}}{\int }\nabla u\nabla \phi \text{d}x+\underset{{ℝ}^{2}}{\int }V\left(x\right)u\phi \text{d}x$ (18)

$\begin{array}{l}〈{I}^{\prime }\left({u}_{n}\right),\phi 〉=\underset{{ℝ}^{2}}{\int }\nabla {u}_{n}\nabla \phi \text{d}x+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}_{n}\phi \text{d}x+\gamma \underset{{ℝ}^{2}}{\int }\theta \left[{u}_{n}\right]{u}_{n}\phi \text{d}x-\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right)\phi \text{d}x\\ \to \underset{{ℝ}^{2}}{\int }\nabla u\nabla \phi \text{d}x+\underset{{ℝ}^{2}}{\int }V\left(x\right)u\phi \text{d}x+\gamma \underset{{ℝ}^{2}}{\int }\theta \left[u\right]u\phi \text{d}x-\underset{{ℝ}^{2}}{\int }g\left(u\right)\phi \text{d}x\end{array}$ (19)

$\begin{array}{l}〈{I}^{\prime }\left({u}_{n}\right),{u}_{n}-u〉=\underset{{ℝ}^{2}}{\int }\nabla {u}_{n}\nabla \left({u}_{n}-u\right)\text{d}x+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}_{n}\left({u}_{n}-u\right)\text{d}x\\ \text{}+\gamma \underset{{ℝ}^{2}}{\int }\theta \left[{u}_{n}\right]{u}_{n}\left({u}_{n}-u\right)\text{d}x-\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right)\left({u}_{n}-u\right)\text{d}x\to 0\end{array}$

$\underset{{ℝ}^{2}}{\int }{|\nabla \left({u}_{n}-u\right)|}^{2}\text{d}x+\underset{{ℝ}^{2}}{\int }V\left(x\right){\left({u}_{n}-u\right)}^{2}\text{d}x\to 0$

${‖{u}_{n}-u‖}_{E}^{2}\to 0$。因此，在E中， ${u}_{n}\to u$

5. 定理1.3的证明

$\left\{\begin{array}{ll}-\Delta u+V\left(x\right)u-\gamma u\theta =\gamma g\left(u\right),\hfill & x\in {ℝ}^{2}.\hfill \\ -\Delta \theta +{c}^{2}\theta ={u}^{2},\hfill & x\in {ℝ}^{2}.\hfill \end{array}$ (20)

$J\left(u,\theta \right)={J}_{1}\left(u\right)-\gamma {J}_{2}\left(u,\theta \right)$ ,

${J}_{1}\left(u\right)=\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}^{2}\right)\text{d}x\text{d}y$ (21)

${J}_{2}\left(u,\theta \right)=-\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla \theta |}^{2}+{c}^{2}{\theta }^{2}\right)\text{d}x+\frac{1}{2}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \text{d}x+\underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x$

$-\Delta u+V\left(x\right)u=\gamma u\theta \left[u\right]+\gamma g\left(u\right),\text{}x\in {ℝ}^{2}$ , (22)

$\begin{array}{c}ΙΙ\left(u\right)=J\left(u,\theta \left[u\right]\right)\\ =\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}^{2}\right)\text{d}x-\frac{\gamma }{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla \theta |}^{2}+{c}^{2}{\theta }^{2}\right)\text{d}x\\ \text{}+\frac{\gamma }{2}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \text{d}x-\gamma \underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x\\ =\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}^{2}\right)\text{d}x-\frac{\gamma }{4}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x-\gamma \underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x\end{array}$ (23)

$〈Ι{Ι}^{\prime }\left(u\right),v〉=\underset{{ℝ}^{2}}{\int }\left(\nabla u\nabla v+V\left(x\right)uv\right)\text{d}x-\gamma \underset{{ℝ}_{2}}{\int }u\theta \left[u\right]v\text{d}x-\gamma \underset{{ℝ}_{2}}{\int }g\left(u\right)v\text{d}x$ ,

$F\left(u\right)={J}_{2}\left(u,\theta \left[u\right]\right)=-\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla \theta \left[u\right]|}^{2}+{c}^{2}{|\theta \left[u\right]|}^{2}\right)\text{d}x+\frac{1}{2}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x+\underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x$

(B) 对于 $\sigma >0$，定义集合 ${M}_{\sigma }=\left\{u\in E|F\left(u\right)=\sigma \right\}$，那么 ${M}_{\sigma }$ 是维数为1的非空 ${C}^{1}$ 流形。

$\begin{array}{c}〈{F}^{\prime }\left(u\right),v〉=\frac{\partial {J}_{2}\left(u,\theta \left[u\right]\right)}{\partial u}v+\frac{\partial {J}_{2}\left(u,\theta \left[u\right]\right)}{\partial \theta }{\theta }^{\prime }\left[u\right]v=\frac{\partial {J}_{2}\left(u,\theta \left[u\right]\right)}{\partial u}v\\ =\underset{{ℝ}^{2}}{\int }u\theta \left[u\right]v\text{d}x+\underset{{ℝ}^{2}}{\int }g\left(u\right)v\text{d}x\end{array}$ (24)

$F\left(u\right)=\frac{1}{4}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x+\underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x$ . (25)

${M}_{\sigma }=\left\{u\in E|\frac{1}{4}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x+\underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x=\sigma \right\}$

$K\left(s\right)=F\left(su\right)=\frac{{s}^{4}}{4}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x+\underset{{ℝ}^{2}}{\int }G\left(su\right)\text{d}x$ ,

$K\left({s}_{0}\right)=F\left({s}_{0}u\right)=\sigma$ .

${\stackrel{¯}{\gamma }}_{0}\left(M\right)=\mathrm{sup}\left\{{\gamma }_{0}\left(K\right):K\subset M\right\}\le \infty$ ,

${J}_{1}\left(u\right)=\frac{1}{2}\underset{{ℝ}^{2}}{\int }\left({|\nabla u|}^{2}+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}^{2}\right)\text{d}x\text{d}y\ge 0$ .

$\underset{n\to \infty }{\mathrm{lim}}F\left({u}_{n}\right)=\underset{n\to \infty }{\mathrm{lim}}\frac{1}{4}\underset{{ℝ}^{2}}{\int }{u}_{n}^{2}\theta \left[{u}_{n}\right]\text{d}x+\underset{{ℝ}^{2}}{\int }G\left({u}_{n}\right)\text{d}x=\frac{1}{4}\underset{{ℝ}^{2}}{\int }{u}^{2}\theta \left[u\right]\text{d}x+\underset{{ℝ}^{2}}{\int }G\left(u\right)\text{d}x=F\left(u\right)$ ,

(i) $\left\{{J}_{1}\left({u}_{n}\right)\right\}$ 有界；

(ii) 当 $n\to \infty$ 时， ${{{J}^{\prime }}_{1}|}_{{M}_{\sigma }}\left({u}_{n}\right)\to 0$

$\underset{{ℝ}^{2}}{\int }\left(\nabla {u}_{n}\nabla v+V\left(x\right){u}_{n}v\right)\text{d}x-{\beta }_{n}\left(\underset{{ℝ}^{2}}{\int }{u}_{n}\theta \left[{u}_{n}\right]v\text{d}x+\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right)v\text{d}x\right)\to 0$ (26)

${J}_{1}\left({u}_{n}\right)=\underset{{ℝ}^{2}}{\int }\left({|\nabla {u}_{n}|}^{2}+\underset{{ℝ}^{2}}{\int }V\left(x\right){u}_{n}^{2}\right)\text{d}x\to d$ . (27)

$\underset{{ℝ}^{2}}{\int }\left({|\nabla {u}_{n}|}^{2}+V\left(x\right){u}_{n}^{2}\right)\text{d}x-{\beta }_{n}\left(\underset{{ℝ}^{2}}{\int }{u}_{n}^{2}\theta \left[{u}_{n}\right]v\text{d}x+\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right){u}_{n}\text{d}x\right)\to 0,\text{}n\to \infty$ (28)

$\underset{{ℝ}^{2}}{\int }{u}_{n}^{2}\theta \left[{u}_{n}\right]\text{d}x+\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right){u}_{n}\text{d}x\ge \underset{{ℝ}^{2}}{\int }{u}_{n}^{2}\theta \left[{u}_{n}\right]\text{d}x+4\underset{{ℝ}^{2}}{\int }G\left({u}_{n}\right)\text{d}x=4\sigma >0$ (29)

$0<\beta =\underset{n\to \infty }{\mathrm{lim}}{\beta }_{n}\le \frac{d}{4\sigma }$ .

$\underset{{ℝ}^{2}}{\int }\left(\nabla {u}_{n}\nabla \left({u}_{n}-u\right)+V\left(x\right){u}_{n}\left({u}_{n}-u\right)\right)\text{d}x-{\beta }_{n}\left(\underset{{ℝ}^{2}}{\int }{u}_{n}\theta \left[{u}_{n}\right]\left({u}_{n}-u\right)\text{d}x+\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right)\left({u}_{n}-u\right)\text{d}x\right)\to 0$ (30)

${S}^{m-1}$${H}_{m}$ 中的单位群。定义映射 $\eta :{S}^{m-1}\to {K}_{m}$$\eta \left(u\right)={\lambda }^{*}\left(u\right)u$，其中 ${\lambda }^{*}\left(u\right)u$ 是直线 $\left\{\lambda u|\lambda >0\right\}$${M}_{\sigma }$ 相交的唯一相交的点。因为 ${M}_{\sigma }$ 是维数为1的非空流形， $\eta$${S}^{m-1}$ 上有定义。因此， $\eta$ 是奇连续映射。由 [15] 的命题5.2，我们有 ${\stackrel{¯}{\gamma }}_{0}\left({S}^{m-1}\right)=m$。因此可得 ${\stackrel{¯}{\gamma }}_{0}\left({K}_{m}\right)\ge m$。由 ${K}_{m}={H}_{m}\cap {M}_{\sigma }\subset {M}_{\sigma }$，可得 ${\stackrel{¯}{\gamma }}_{0}\left({M}_{\sigma }\right)\ge {\stackrel{¯}{\gamma }}_{0}\left({K}_{m}\right)\ge m$。因为m是任意的，我们有 ${\stackrel{¯}{\gamma }}_{0}\left({M}_{\sigma }\right)=\infty$

$\underset{{ℝ}^{2}}{\int }\left(\nabla {u}_{n}\nabla v+V\left(x\right){u}_{n}v\right)\text{d}x-{\gamma }_{n}\left(\underset{{ℝ}^{2}}{\int }{u}_{n}\theta \left[{u}_{n}\right]v\text{d}x+\underset{{ℝ}^{2}}{\int }g\left({u}_{n}\right)v\text{d}x\right)=0$ ,

6. 结论

NOTES

*第一作者。

#通讯作者。

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